# How do you solve for x in log_2x=log_4(25)?

May 1, 2018

$x = 5$

#### Explanation:

Using the formal definiton of a logarithm, we can take the right hand side to be:

${\log}_{4} 25 = a \implies {4}^{a} = 25$

Since $4$ is simply ${2}^{2}$, we have:

${2}^{2 a} = 25$

Remember; $a$ is equal to ${\log}_{4} 25$, which is in turn equal to ${\log}_{2} x$.

${2}^{2 {\log}_{2} x} = 25$

Taking the binary logarithm of both sides:

$2 {\log}_{2} x = {\log}_{2} 25$

$\implies {\log}_{2} x = \frac{1}{2} {\log}_{2} 25$

Knowing that $a {\log}_{b} c = {\log}_{b} {c}^{a}$, we reach the result we wished to get:

${\log}_{2} x = {\log}_{2} {\left(25\right)}^{\frac{1}{2}} = {\log}_{2} 5 \implies \textcolor{red}{x = 5}$